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Grade 11General Physics

the motion of a particle in a straight line is defined by the relation x=t^4-12t^2-40

where x is in metres and t is in sec.determine the position x,velocity v, and acceleration

a of the particle at t=2sec

Profile image of prince parihar
12 Years agoGrade 11
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3 Answers

Profile image of Khandavalli Satya Srikanth
12 Years ago

At t=2sec position x=(2^4-12*2^2-40)=-80 (-ve sign indicates particle moving in -ve x direction)

velocity v=dx/dt=4*t^3-24*t

at t=2sec v=4*2^3-24*2=-16m/sec

acceleration=dv/dt=12*t^2-24

at t=2sec a=12*2^2-24=24m/sec^2

Profile image of girijesh tripathi
12 Years ago

posiposition-68m    v--     -16m/s   accl--- + 24m/s2

Profile image of pranjal sharma
12 Years ago

velocity= -80m/s

accleration = - 8m/s2

posiion= -72